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Logical solving path
   LMI Essentials -> Solving Techniques3 posts • Page 1 of 1 • 1
vopani
Subject: Logical solving path @ 2015-10-21 2:07 PM (#19647) (#19647) Top


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Hi, can someone solve this Akari puzzle logically?
Puzzle taken from here: http://wa1729.blogspot.in/2015/10/puzzle-59-akari.html





(Akari.png)



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prasanna16391
Subject: Re: Logical solving path @ 2015-10-21 7:10 PM (#19649 - in reply to #19647) (#19649) Top


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Here's what I did (there's a little bit of forward thinking but I could visualize it easily enough) -

I started by marking some of the 'L' triminoes which must have a light. Two such triminoes are R1C5, R1C6, R2C6 and R2C1, R3C1, R3C2.

In the latter, either R2C1 or R3C2 must be a light.
If its R2C1, then using the 1st Trimino, R1C5 must be a light, which forms another L trimino in R1C2, R1C3 and R2C3 where R1C2 must be a light.

Using the above two statements, the rest of C2 can be marked out since either R1C2 or R3C2 must be a light.

Now extend the 'L' thinking to larger Ls. For the L from R5C1 to R6C5 via R6C1, either of R6C1, C3 or C4 must be a light. Similarly, for the L from R9C1-R8C1-R8C4, either of R8C1, C3 or C4 must be a light. When you put these two statements together with the fact that R6C1+R8C1= 1 light, the other four cells must also have 1 light, and R9C4 and R10C3 can have only one light between them for the 2 in R10C4. You have a defined area for 2 lights in C3 and C4, so the rest can be marked as otherwise.

There are a few more deductions after this, because R6C1 will need to be a light to illuminate R5C1, and if you remember the above reasoning, R2C1 containing a light means R1C2 must contain a light and you will see at this stage that it is not possible.

I'm stuck too after all this :P
prasanna16391
Subject: Re: Logical solving path @ 2015-10-21 7:19 PM (#19650 - in reply to #19647) (#19650) Top


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Ok got something more from noticing that R3C7 being a light basically forces some lights in the bottom right which make it impossible to place a light in the L from R4C6-R5C6-R5C10. This basically means R3C5 is a light to illuminate R1C5. Then using the fact that one light will be needed in R4 to eliminate R4C6, along with the 2 clue in C8, it is possible to fix an area where 3 lights appear in C7, C8, C9. From here it is easy to finish.
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